Question: Simplify the following expression and state the condition under which the simplification is valid: $r = \dfrac{y^2 - 4y}{y^2 - 8y + 16}$
Answer: First factor the expressions in the numerator and denominator. $ \dfrac{y^2 - 4y}{y^2 - 8y + 16} = \dfrac{(y)(y - 4)}{(y - 4)(y - 4)} $ Notice that the term $(y - 4)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(y - 4)$ gives: $r = \dfrac{y}{y - 4}$ Since we divided by $(y - 4)$, $y \neq 4$. $r = \dfrac{y}{y - 4}; \space y \neq 4$